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\begin{document}

\section*{\huge Additional notes for quantitative results}

\section{Calibration algorithm}

Disutility from labour is given by $h(\ell) = \frac{\ell^{1 + \frac{1}{\varepsilon}}}{1 + \frac{1}{\varepsilon}}$, which is common to all workers. We reintroduce $\pi(\ell)$ function, which is a probability of high output realisation at performance pay jobs. We suppose it takes a power function form $\pi(\ell) = \bar\pi \ell^{\rho}$, $\rho\in(0, 1]$, with the inverse function $\pi^{-1}(x) = (\frac{x}{\bar\pi})^{\frac{1}{\rho}}$. Define auxiliary disutility from effort expressed as probability of high output $h_{pp}(x) = \bar h_{pp} \frac{x^{1 + \frac{1}{\varepsilon_{pp}}}}{1 + \frac{1}{\varepsilon_{pp}}}$, where $h_{pp}(x) = h(\pi^{-1}(x))$.\footnote{It is easy to show that $\varepsilon_{pp} = \frac{\rho\varepsilon}{1 + \varepsilon - \rho\varepsilon}$ and $\bar h_{pp} = \bar \pi^{-\frac{1 + \varepsilon}{\rho\varepsilon}}\frac{1 + \frac{1}{\varepsilon_{pp}}}{1 + \frac{1}{\varepsilon}}$.} Denote the variance of log-earnings risk at performance-pay jobs $Var (\log z_{pp}\mid\theta) = \beta^2(1 - \pi(l_{pp}))\pi(l_{pp})$ by $\sigma^2_{\beta}$. 

We match the following moments and assumptions:
\begin{enumerate}
  \item Ratio of variance of log-earnings at the two job types: $m_1 = \frac{\lambda_{\theta}^{-2} + \sigma^2_{\theta, pp} + \sigma^2_{\beta}}{\lambda_{\theta}^{-2} + \sigma^2_{\theta, fp}}$
  \item Assumption: risky bonuses explain share $m_2 > 0$ of excess variance of log-earnings at pp jobs: $m_2 = \frac{\sigma^2_{\beta}}{\sigma^2_{\beta} + \sigma^2_{\theta, pp} - \sigma^2_{\theta,fp}} = m_2 \implies \sigma^2_{\theta,pp} = \frac{1 - m_2}{m_2}\sigma^2_\beta + \sigma^2_{\theta,fp}$.
  \item Ratio of mean earnings at the two job types: $ m_3 = \frac{\mathbb{E}[\theta_{pp}]\pi}{\mathbb{E}[\theta_{fp}]\ell_{fp}}$.
  \item Cross-sectional variance of log-earnings $m_4 = Var(\log z)$. 
\end{enumerate}

The calibration algorithm proceeds as follows:
\begin{enumerate}
  \item Guess value of $\sigma^2_{fp}$.
  \item Calculate $\sigma^2_{\beta} = m_2(m_1 - 1)(\lambda_{\theta}^{-2} + \sigma^2_{\theta, fp})$ (follows from moments 1 and 2)
  \item Calculate $\beta = \sqrt{\frac{\sigma^2_{\beta}}{\pi(1 - \pi)}}$ (from the definition of $\sigma^2_{\beta}$)
  \item Calculate $\varepsilon_{pp} = \sigma^2_{\beta} \frac{b}{\underline z \beta} = \sigma^2_{\beta} \frac{e^{\beta} - 1}{\beta}$ (from the effort FOC)
  \item Calculate $\rho = \frac{1 + \varepsilon}{\varepsilon}\frac{\varepsilon_{pp}}{1 + \varepsilon_{pp}}$ (by inverting $\varepsilon_{pp} = \frac{\rho\varepsilon}{1 + \varepsilon - \rho\varepsilon}$ from  footnote above)
  \item Calculate $\bar h_{pp} = \bar \pi^{-\frac{1 + \varepsilon}{\rho\varepsilon}}\frac{1 + \frac{1}{\varepsilon_{pp}}}{1 + \frac{1}{\varepsilon}}$ (from the footnote above)
  \item Given $\rho$ and $h_{pp}$, calculate $\bar\pi$
  \item Calculate $\ell_{fp} = (1 - p)^{\frac{\varepsilon}{1 + \varepsilon}}$.
  \item Calculate $\mu_{pp} - \mu_{fp}$ by matching $m_3$
  \begin{align}
    m_3  & = \frac{\mathbb{E}[\theta_{pp}]}{\mathbb{E}[\theta_{fp}]}\frac{\pi}{\ell_{fp}} = e^{\mu_{pp} - \mu_{fp} + \frac{\sigma^2_{\theta, pp} - \sigma^2_{\theta, fp}}{2}} \frac{\pi}{\ell_{fp}} = e^{\mu_{pp} - \mu_{fp} + \frac{1 - m_2}{m_2}\frac{\sigma^2_{\beta}}{2}} \frac{\pi}{\ell_{fp}} \\
    & \implies \mu_{pp} - \mu_{fp} = \log\left(m_3 \frac{\ell_{fp}}{\pi}\right) - \frac{1 - m_2}{m_2}\frac{\sigma^2_{\beta}}{2}
  \end{align}

  \item Calculate base pay of performance-pay worker with $\theta =1$: $\underline z_{pp}(1) = \frac{\pi}{1 - \pi + \pi e^{\beta}}$.
  \item Calculate cross-sectional variance of log earnings
  \begin{align*}
    Var(\log z) 
    & = \left(m_1 s_{pp} + 1 - s_{pp}\right) (\lambda_{\theta}^{-2} + \sigma^2_{\theta, fp}) \\
    & + s_{pp} (1 - s_{pp}) \left(\log \left(e^{\mu_{pp}}\cdot\underline z_{pp}(1)\right) + \pi\beta - \log(e^{\mu_{fp}}\cdot \ell_{fp})\right)^2. 
  \end{align*}
  It is does not equal $m_4$, adjust the initial guess of $\sigma^2_{fp}$.
\end{enumerate}

\section{Analytical formula for $\mathcal{E}^*$ for computation of top tax rates}

Denote the elasticity of labour effort of type $\theta$ with respect to the top tax rate by $\varepsilon_{\ell, 1-\tau}(\theta) = \frac{\hat \ell(\theta)}{-\hat \tau}\frac{1 - \tau}{\ell(\theta)}$. Otherwise, notation is as in the proof of Proposition 3 in the Online Appendix. The average elasticity among top earners, weighted by expected earnings, is equal
\begin{align}
  \mathcal{E} & = \underbrace{\int_{\theta^{*}}^{\theta^{**}}\frac{\theta\ell}{\bar Z} \varepsilon_{\ell, 1 - \tau}\frac{dF_{\theta}(\theta)}{1 - F_{\theta}(\theta^*)}}_{=e_1} + \underbrace{\int_{\theta^{**}}^{\infty}\frac{\theta\ell}{\bar Z} \varepsilon_{\ell, 1 - \tau}\frac{dF_{\theta}(\theta)}{1 - F_{\theta}(\theta^*)}}_{=e_2}.
  % & \overset{MVT}{=} \varepsilon_{\ell, 1 - \tau}(t_1) \int_{\theta^{*}}^{\theta^{**}}\frac{\theta\ell}{\bar Z} \frac{dF_{\theta}(\theta)}{1 - F_{\theta}(\theta^*)} + \varepsilon_{\ell, 1 - \tau}(t_2)\int_{\theta^{**}}^{\infty}\frac{\theta\ell}{\bar Z} \frac{dF_{\theta}(\theta)}{1 - F_{\theta}(\theta^*)}
\end{align}
% for some $t_1\in(\theta^{*},\theta^{**})$ and $t_2 > \theta^{**}$.

\textbf{Calculating $e_1$.} It follows from Lemma 9 in the Online Appendix that for $\theta\in(\theta^*, \theta^{**})$
\begin{equation}
  \varepsilon_{\ell,1-\tau}(\theta) = X\left(\frac{1 + \tilde R_0/\bar z}{1 - \underline z/\bar z} - \frac{\tilde R_0/\bar z}{\ell + (1 - \ell)\underline z/\bar z + \tilde R_0/\bar z} \frac{\bar z - z^*}{\bar z}\right) \underset{z^*\to\infty}{\to} \frac{X^*}{1 - e^{-h'(\ell^*)}}
\end{equation}
where $\tilde R_0 = \frac{R_0}{1-\tau}$ and $X(\theta) = \left(-\frac{\ell(\theta)^2}{\underline z}\frac{\partial^2\Pi}{\partial\ell^2}\right)^{-1}$ with limit $X^*$ as $z^*\to\infty$ (see Lemma 9 in the Online Appendix for the exact expression).
Then, for some $t_1\in(\theta^*, \theta^{**})$ and $t_2 > \theta^{**}$, we have
\begin{align}
  e_1 & = \frac{\int_{\theta^{*}}^{\theta^{**}}\theta\ell \varepsilon_{\ell, 1 - \tau}dF_{\theta}(\theta)}{\int_{\theta^{*}}^{\infty}\theta\ell dF_{\theta}(\theta)} \overset{MVT}{=} \varepsilon_{\ell, 1 - \tau}(t_1)\frac{\ell(t_1)}{\ell(t_2)} \frac{\int_{\theta^{*}}^{\theta^{**}}\theta dF_{\theta}(\theta)}{\int_{\theta^{*}}^{\infty}\theta dF_{\theta}(\theta)} \\ & = \varepsilon_{\ell, 1 - \tau}(t_1)\frac{\ell(t_1)}{\ell(t_2)} \left(1 - \left(\frac{\theta^*}{\theta^{**}}\right)^{\rho^* - 1}\right) \underset{z^*\to\infty}{\to} \frac{X^*}{1 - e^{-h'(\ell^*)}} \left(1 -  e^{-(\rho^* - 1)h'(\ell^*)}\right)
\end{align}
\textbf{Calculating $e_2$.} It follows from Lemma 9 in the Online Appendix that for $\theta > \theta^{**}$
\begin{align}
  \varepsilon_{\ell,1-\tau}(\theta) & = X\left(1 
  - \frac{\ell \bar z/\underline z + 1 - \ell + \ell\tilde R_0/\bar z}{\ell \bar z/\bar z + 1 - \ell + \tilde R_0/\bar z}\cdot\frac{\underline z - z^*}{\underline z}
  - \frac{\tilde R_0/\bar z}{\ell + (1 - \ell)\underline z/\bar z + \tilde R_0/\bar z} \frac{\bar z - z^*}{\bar z}\right). 
  % \\
  % & = X\left(\frac{\ell \bar z/\underline z + 1 - \ell + \ell\tilde R_0/\bar z}{\ell \bar z/\bar z + 1 - \ell + \tilde R_0/\bar z}\cdot\frac{z^*}{\underline z} - \frac{(1 - \ell)\tilde R_0/\bar z}{\ell \bar z/\bar z + 1 - \ell + \tilde R_0/\bar z}
  % - \frac{\tilde R_0/\bar z}{\ell + (1 - \ell)\underline z/\bar z + \tilde R_0/\bar z} \frac{\bar z - z^*}{\bar z}\right).
\end{align}
Ignoring the terms which vanish as $z^*\to\infty$, we can write
\begin{equation}
  \varepsilon_{\ell,1-\tau}(\theta) \sim X \frac{\ell \bar z/\underline z + 1 - \ell + \ell\tilde R_0/\bar z}{\ell \bar z/\bar z + 1 - \ell + \tilde R_0/\bar z}\cdot\frac{z^*}{\underline z} \equiv \tilde \varepsilon_{\ell,1-\tau}(\theta).
\end{equation}
By a straightforward application of MVT, any terms in $\varepsilon_{\ell,1-\tau}(\theta)$ which vanish in the limit have no impact $\lim_{z^*\to\infty} e_2$. Thus, when calculating the limit of $e_2$, we can replace $\varepsilon_{\ell,1-\tau}(\theta)$ with $\tilde\varepsilon_{\ell,1-\tau}(\theta)$. Proceeding in this way, we obtain
\begin{align}
  \lim_{z^*\to\infty} e_2 &
  = \lim_{z^*\to\infty} \frac{\int_{\theta^{**}}^{\infty}\theta\ell \tilde\varepsilon_{\ell, 1 - \tau}dF_{\theta}(\theta)}{\int_{\theta^{*}}^{\infty}\theta\ell dF_{\theta}(\theta)} 
  % = \lim_{z^*\to\infty} \frac{\theta^{**}\int_{\theta^{**}}^{\infty}\ell \frac{\theta}{\theta^{**}}\tilde\varepsilon_{\ell, 1 - \tau}dF_{\theta}(\theta)}{\int_{\theta^{*}}^{\infty}\theta\ell dF_{\theta}(\theta)} \\
  \overset{MVT}{=} \lim_{z^*\to\infty} \frac{\ell(t_1)}{\ell(t_2)} \frac{t_1}{\theta^{**}}\tilde\varepsilon_{\ell, 1 - \tau}(t_1) \frac{\theta^{**}\int_{\theta^{**}}^{\infty}dF_{\theta}(\theta)}{\int_{\theta^{*}}^{\infty}\theta dF_{\theta}(\theta)}.
\end{align}
Plug in the expression for $\tilde\varepsilon_{\ell, 1 - \tau}(t_1)$ and use $\frac{t_1}{\theta^{**}}\frac{z^*}{\underline{z}(t_1)}\to 1$ as $z^*\to\infty$, as well as the properties of the Pareto distribution, to get
\begin{align}
  \lim_{z^*\to\infty} e_2 &
  = \lim_{z^*\to\infty} X(t_1) \frac{\rho^* - 1}{\rho^*} 
  \left(\frac{\theta^*}{\theta^{**}}\right)^{\rho^* - 1} = X^* \frac{\rho^* - 1}{\rho^*} e^{-(\rho^* - 1)h'(\ell^*)}.
\end{align}
Summing up, we have 
\begin{align}
  \mathcal{E}^* & = \lim_{z^* \to\infty} [e_1 + e_2] = \frac{X^*}{1 - e^{-h'(\ell^*)}} \left(1 -  e^{-(\rho^* - 1)h'(\ell^*)}\right) + X^* \frac{\rho^* - 1}{\rho^*} e^{-(\rho^* - 1)h'(\ell^*)}.
\end{align}

\end{document}